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  • Using the properties of determinants in prove that: a^2+2a,2a+1,1

Using the properties of determinants, prove that:
\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|=(a-1)^{3}

Answers (1)

Take LHS

\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|

\\\begin{array}{l} \text { Apply, } R_{1} \rightarrow R_{1}-R_{2} \\ =\left|\begin{array}{ccc} a^{2}+2 a-(2 a+1) & 2 a+1-(a+2) & 1-1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \\ \\=\left|\begin{array}{ccc} a^{2}+2 a-2 a-1 & 2 a+1-a-2 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \\ \\=\left|\begin{array}{ccc} a^{2}-1 & a-1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \end{array}

\\\begin{aligned} &=\left|\begin{array}{ccc} (a-1)(a+1) & a-1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|\left[\because\left(a^{2}-b^{2}\right)=(a-b)(a+b)\right]\\ &\text { Take (a-1) common from R }\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|\\ &\text { Apply, } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a+1-3 & a+2-3 & 1-1 \\ 3 & 3 & 1 \end{array} \mid\right|.\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1 \end{array}\right|\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1 \end{array}\right| \end{aligned}

Take (a-1) common from R_2

\begin{aligned} &=(a-1)^{2}\left|\begin{array}{ccc} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right|\\ &\text { Expand along } \mathrm{C}_{3} \end{aligned}

\\\\ = (a -1)\textsuperscript{2} [1$ \{ $ (a + 1) -2$ \} $ ]\\ \\ \vspace{\baselineskip}= (a -1)\textsuperscript{2} [a + 1 -2]\\ \\ \vspace{\baselineskip}= (a -1)\textsuperscript{3}\\ \\ \vspace{\baselineskip}= RHS\\

Hence, proved.

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infoexpert22

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